Answer:
66.28 s
Explanation:
Step 1:
Acceleration,a=[tex]0.065m/s^2[/tex]
Time,t=9.75 min=[tex]9.75\times 60=585s[/tex]
1 min=60 s
Initial velocity,u=[tex]3.4m/s[/tex]
We have to find the final velocity of train.
We know that
[tex]v=u+at[/tex]
Substitute the values
[tex]v=3.4+0.065(585)=41.425m/s[/tex]
Step 2:
Now, initial velocity,u=41.425m/s
Deceleration,a=[tex]-0.625m/s^2[/tex]
Because the train velocity decreases
Final velocity, v=0
Again, substitute the values in the above formula
[tex]0-41.425=-0.625t[/tex]
[tex]-41.425=-0.625t[/tex]
[tex]t=\frac{-41.425}{-0.625}[/tex]
[tex]t=66.28s[/tex]
Hence, the train takes 66.28 s to come to stop from velocity 41.425m/s.
